I do not understand how the fact $M$ is maximal is related to $A/M$ being simple.
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- $\begingroup$ If $I/M$ is a proper non-trivial ideal of $A/M$, what can you say about $\langle I, M \rangle$? $\endgroup$Robert Shore– Robert Shore2023-05-03 02:29:49 +00:00Commented May 3, 2023 at 2:29
- $\begingroup$ Provided $M$ is maximal, I can assure $\langle I, M\rangle = A$. Is that right? $\endgroup$user1120269– user11202692023-05-03 02:31:21 +00:00Commented May 3, 2023 at 2:31
- 1$\begingroup$ Hint: Think of the correspondence isomorphism theorems. Specifically every proper ideal $J \leq A/M$ is of the form $I/M$ for some proper ideal $I$ of $A$ which contains $M$. $\endgroup$Irving Rabin– Irving Rabin2023-05-03 02:33:24 +00:00Commented May 3, 2023 at 2:33
- $\begingroup$ And can that be possible if $I/M$ is a proper, non-trivial ideal of $A/M$? $\endgroup$Robert Shore– Robert Shore2023-05-03 02:42:29 +00:00Commented May 3, 2023 at 2:42
- $\begingroup$ No, we previously stated $I/M$ is non trivial and because of that $I \neq M$. Also, $I \neq A$ because $I/M$ is a proper ideal. Contradiction. $\endgroup$user1120269– user11202692023-05-03 02:51:07 +00:00Commented May 3, 2023 at 2:51
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